Given two integers n and m, in a single operation n can be multiplied by either 2 or 3. The task is to convert n to m with a minimum number of given operations. If it is impossible to convert n to m with the given operation then print -1.
Examples:
Input: n = 120, m = 51840
Output: 7
120 * 2 * 2 * 2 * 2 * 3 * 3 * 3 = 51840
Input: n = 42, m = 42
Output: 0
No operation required.
Input: n = 48, m = 72
Output: -1
Approach: If m is not divisible by n then print -1 as n cannot be converted to m with the given operation. Else we can check if, on dividing, the quotient has only 2 and 3 as prime factors. If yes then the result will be the sum of powers of 2 and 3 else print -1
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the minimum// operations requiredint minOperations(int n, int m){ if (m % n != 0) return -1; int minOperations = 0; int q = m / n; // Counting all 2s while (q % 2 == 0) { q = q / 2; minOperations++; } // Counting all 3s while (q % 3 == 0) { q = q / 3; minOperations++; } // If q contained only 2 and 3 // as the only prime factors // then it must be 1 now if (q == 1) return minOperations; return -1;}// Driver codeint main(){ int n = 120, m = 51840; cout << minOperations(n, m); return 0;} |
Java
// Java implementation of the approachclass GfG { // Function to return the minimum // operations required static int minOperations(int n, int m) { if (m % n != 0) return -1; int minOperations = 0; int q = m / n; // Counting all 2s while (q % 2 == 0) { q = q / 2; minOperations++; } // Counting all 3s while (q % 3 == 0) { q = q / 3; minOperations++; } // If q contained only 2 and 3 // as the only prime factors // then it must be 1 now if (q == 1) return minOperations; return -1; } // Driver code public static void main(String[] args) { int n = 120, m = 51840; System.out.println(minOperations(n, m)); }} |
Python3
# Python 3 implementation of the approach# Function to return the minimum# operations requireddef minOperations(n, m): if (m % n != 0): return -1 minOperations = 0 q = int(m / n) # Counting all 2s while (q % 2 == 0): q = int(q / 2) minOperations += 1 # Counting all 3s while (q % 3 == 0): q = int(q / 3) minOperations += 1 # If q contained only 2 and 3 # as the only prime factors # then it must be 1 now if (q == 1): return minOperations return -1# Driver codeif __name__ == '__main__': n = 120 m = 51840 print(minOperations(n, m)) # This code is contributed by# Surendra_Gangwar |
C#
// C# implementation of the approachusing System;class GFG {// Function to return the minimum// operations requiredstatic int minOperations(int n, int m){ if (m % n != 0) return -1; int minOperations = 0; int q = m / n; // Counting all 2s while (q % 2 == 0) { q = q / 2; minOperations++; } // Counting all 3s while (q % 3 == 0) { q = q / 3; minOperations++; } // If q contained only 2 and 3 // as the only prime factors // then it must be 1 now if (q == 1) return minOperations; return -1;}// Driver codepublic static void Main(){ int n = 120, m = 51840; Console.WriteLine(minOperations(n, m));}}// This code is contributed // by Akanksha Rai |
PHP
<?php// PHP implementation of the approach// Function to return the minimum// operations requiredfunction minOperations($n, $m){ if ($m % $n != 0) return -1; $minOperations = 0; $q = $m / $n; // Counting all 2s while ($q % 2 == 0) { $q = $q / 2; $minOperations++; } // Counting all 3s while ($q % 3 == 0) { $q = $q / 3; $minOperations++; } // If q contained only 2 and 3 // as the only prime factors // then it must be 1 now if ($q == 1) return $minOperations; return -1;}// Driver code$n = 120; $m = 51840;echo(minOperations($n, $m));// This code is contributed by Code_Mech?> |
Javascript
<script>// javascript implementation of the approach // Function to return the minimum // operations required function minOperations(n , m) { if (m % n != 0) return -1; var minOperations = 0; var q = m / n; // Counting all 2s while (q % 2 == 0) { q = q / 2; minOperations++; } // Counting all 3s while (q % 3 == 0) { q = q / 3; minOperations++; } // If q contained only 2 and 3 // as the only prime factors // then it must be 1 now if (q == 1) return minOperations; return -1; } // Driver code var n = 120, m = 51840; document.write(minOperations(n, m));// This code contributed by Rajput-Ji </script> |
Output:
7
Time Complexity: O(log(m/n))
Auxiliary Space: O(1)
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